Z Complex Minima
If Z is a non-zero complex number, then find the minimum value of
( Im ( Z ) ) 5 Im ( Z 5 ) .
Clarification: Im ( Z ) represent the imaginary part of Z .
The answer is -4.
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3 solutions
The solution can be completely elementary if you completed squares rather than differentiation. Especially considering a lot of younger users in this place, such a solution will be completely comprehensive. Anyway, nice work!
I am talking about writing 5 w 4 − 1 0 w 2 + 1 = 5 ( w 2 − 1 ) 2 − 4 ≥ − 4 .
It would be helpful to first mention that [ I m ( x + y i ) ] 5 = y 5 , and likewise for I m [ ( x + y i ) 5 ] . Otherwise, people might be confused in understanding the equation, since [ I m ( x + y i ) ] 5 = y 4 .
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I've edited it to make it a bit clearer.
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Perfect :)
to be honest, it tripped me up when I first read it, which is why I made the comment.
Nice! While the initial stages (expanding ( x + i y ) 5 ) were pretty nasty, the derivative of f ( w ) looks way easier than differentiating sin 5 x sin ( 5 x ) . Also, you forgot to mention that while the solution w = 0 gives a maximum, two solutions w = ± 1 both give minima.
Nice derivation, a couple minor corrections:
- Second term of the second equation should be negative, same for the following equations. This is corrected at the w = x / y substitution.
- There is a third solution to the derivative: w = − 1 . This also evaluates to f ( − 1 ) = − 4 though, so it doesn't affect the answer.
WLOG let Z = e i x = cos x + i sin x . Hence, I m ( Z 5 ) = I m ( e 5 i x ) = sin ( 5 x ) and I m 5 ( Z ) = sin 5 x .
We find d x d sin 5 x sin ( 5 x ) = − 1 0 csc 5 x ( cos x + cos ( 3 x ) ) , eliminate non-zero terms and set it to 0:
cos x + cos ( 3 x ) = 0
We can see that the roots occur at x i = ± 4 π , ± 2 π , ± 4 3 π , ± 4 5 π , … ; checking D ( x i ) = d x d cos x + cos ( 3 x ) ∣ x = x i shows that every other root, ie x 2 i = − 2 π , 4 π , 4 3 π , … is a maximum since D ( x 2 i ) < 0 .
Substituting in any x 2 i into sin 5 x sin ( 5 x ) gives
min ( sin 5 x sin ( 5 x ) ) = ( 2 1 ) 5 − 2 1 = − 4
That's the better way ! (although ultimately same ) Good Work ! :D
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I think Adam's method is slightly better because it's purely algebraic (other than differentiation).
I have one doubt in Jake's method. It restrict the Re(Z) and Im(Z) between -1 and 1 as the coefficients are sin and cos function which can't lie outside -1 and 1. This kind of restriction is not there in Adam's method
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To make it general write Z = r e i x = r ( c o s x + i s i n x ) Ultimately the answer is same because r cancels out.
let z=x+iy Now, (x+iy)^2=(x^2-y^2 +2xyi)……(1) Again multiply eqn 1 two times We get (x+iy)^4 Now [Im{(x+iy) *(x+iy)^4}/Im(z^5)]=5(x/y)^4 -10(x/y)^2 +1 Let us assume f(y)=5(x/y)^4 -10(x/y)^2 +1 Now df/dy=0 gives [x^2=y^2] Putting above we get the result -4
Change z to x + y i I m ( x + y i ) 5 I m ( ( x + y i ) 5 ) I m ( x + y i ) is just y . Expand brackets in the numerator and take the imaginary part to get the overall expression: y 5 y ( 5 x 4 + 1 0 x 2 y 2 + y 4 ) Simplify: y 4 5 x 4 + 1 0 x 2 y 2 + y 4 5 y 4 x 4 + 1 0 y 2 x 2 + 1 Let w = y x and set the expression to f ( w ) . f ( w ) = 5 w 4 − 1 0 w 2 + 1 Derive the new function and then set the derivative to 0 to find where the minimum occurs: f ′ ( w ) = 2 0 w 3 − 2 0 w 0 = 2 0 w ( w 2 − 1 ) Solving this equation gives w = 0 or w = 1 . Substitute values back into the original function to find the minimum and maximum: f ( 0 ) = 5 ( 0 ) 4 − 1 0 ( 0 ) 2 + 1 = 1 f ( 1 ) = 5 ( 1 ) 4 − 1 0 ( 1 ) 2 + 1 = − 4
The minimum is − 4 as − 4 < 1 .